Set all vector elements to NA in a list of vectors
6
How can I set all vector elements to NA in a list of vectors?
Essentially, I'd like to keep an existing list's structure and names but empty all values, to fill them in later. I provide a minimal example with a couple solutions below. I prefer base and tidyverse (esp. purrr) solutions, but can get on board with any approach which is better than what I have below.
my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))
ret_list <- my_list
# Approach 1
for (element_name in names(my_list)) {
ret_list[[element_name]] <- NA
}
ret_list
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
# Approach 2
lapply(my_list, function(x) {x <- NA; return(x)})
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
r
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
add a comment |
6
How can I set all vector elements to NA in a list of vectors?
Essentially, I'd like to keep an existing list's structure and names but empty all values, to fill them in later. I provide a minimal example with a couple solutions below. I prefer base and tidyverse (esp. purrr) solutions, but can get on board with any approach which is better than what I have below.
my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))
ret_list <- my_list
# Approach 1
for (element_name in names(my_list)) {
ret_list[[element_name]] <- NA
}
ret_list
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
# Approach 2
lapply(my_list, function(x) {x <- NA; return(x)})
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
r
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
1
both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?
– Ben Bolker
Dec 31 '18 at 23:27
1
Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.
– lowndrul
Dec 31 '18 at 23:40
add a comment |
6
6
6
1
How can I set all vector elements to NA in a list of vectors?
Essentially, I'd like to keep an existing list's structure and names but empty all values, to fill them in later. I provide a minimal example with a couple solutions below. I prefer base and tidyverse (esp. purrr) solutions, but can get on board with any approach which is better than what I have below.
my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))
ret_list <- my_list
# Approach 1
for (element_name in names(my_list)) {
ret_list[[element_name]] <- NA
}
ret_list
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
# Approach 2
lapply(my_list, function(x) {x <- NA; return(x)})
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
r
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
How can I set all vector elements to NA in a list of vectors?
Essentially, I'd like to keep an existing list's structure and names but empty all values, to fill them in later. I provide a minimal example with a couple solutions below. I prefer base and tidyverse (esp. purrr) solutions, but can get on board with any approach which is better than what I have below.
my_list <- list(A = c('a' = 1, 'b' = 2, 'c' = 3), B = c('x' = 10, 'y' = 20))
ret_list <- my_list
# Approach 1
for (element_name in names(my_list)) {
ret_list[[element_name]] <- NA
}
ret_list
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
# Approach 2
lapply(my_list, function(x) {x <- NA; return(x)})
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
r
r
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
edited Jan 1 at 0:02
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
asked Dec 31 '18 at 22:56
lowndrul
1,16521835
1,16521835
1
both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?
– Ben Bolker
Dec 31 '18 at 23:27
1
Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.
– lowndrul
Dec 31 '18 at 23:40
add a comment |
1
both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?
– Ben Bolker
Dec 31 '18 at 23:27
1
Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.
– lowndrul
Dec 31 '18 at 23:40
1
1
both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?
– Ben Bolker
Dec 31 '18 at 23:27
both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?
– Ben Bolker
Dec 31 '18 at 23:27
1
1
Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.
– lowndrul
Dec 31 '18 at 23:40
Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.
– lowndrul
Dec 31 '18 at 23:40
add a comment |
5 Answers
5
active
oldest
votes
8
Here's another one for numeric vectors:
lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
More generally,
lapply(my_list, replace, TRUE, NA)
and
lapply(ret_list, ifelse, NA, NA)
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
what is*called as in this context ?
– YOLO
Dec 31 '18 at 23:12
1
@YOLO, I'm not sure what you mean. The first approach is the same aslapply(my_list, function(x) x * NA), so*is the usual product, which results toNAwhen one of the factors isNA.
– Julius Vainora
Dec 31 '18 at 23:15
add a comment |
7
You can use function is.na<- in a lapply loop.
ret_list <- lapply(my_list, `is.na<-`)
ret_list
#$A
# a b c
#NA NA NA
#
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
add a comment |
4
another alternative with dplyr:
lapply(my_list, function(x) dplyr::na_if(x,x))
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
add a comment |
4
If the list is not restricted to one level then use rapply.
# test data modified from question
my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))
rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")
which can also be written as:
rapply(my_list2, replace, list = TRUE, values = NA, how = "list")
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
add a comment |
3
Another way around
relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)
#$A
# a b c
#NA NA NA
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:30
989
8,91251834
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
Here's another one for numeric vectors:
lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
More generally,
lapply(my_list, replace, TRUE, NA)
and
lapply(ret_list, ifelse, NA, NA)
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
what is*called as in this context ?
– YOLO
Dec 31 '18 at 23:12
1
@YOLO, I'm not sure what you mean. The first approach is the same aslapply(my_list, function(x) x * NA), so*is the usual product, which results toNAwhen one of the factors isNA.
– Julius Vainora
Dec 31 '18 at 23:15
add a comment |
8
Here's another one for numeric vectors:
lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
More generally,
lapply(my_list, replace, TRUE, NA)
and
lapply(ret_list, ifelse, NA, NA)
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
what is*called as in this context ?
– YOLO
Dec 31 '18 at 23:12
1
@YOLO, I'm not sure what you mean. The first approach is the same aslapply(my_list, function(x) x * NA), so*is the usual product, which results toNAwhen one of the factors isNA.
– Julius Vainora
Dec 31 '18 at 23:15
add a comment |
8
8
8
Here's another one for numeric vectors:
lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
More generally,
lapply(my_list, replace, TRUE, NA)
and
lapply(ret_list, ifelse, NA, NA)
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
Here's another one for numeric vectors:
lapply(my_list, `*`, NA) # Instead of * it could also be +, -, etc.
# $A
# a b c
# NA NA NA
#
# $B
# x y
# NA NA
More generally,
lapply(my_list, replace, TRUE, NA)
and
lapply(ret_list, ifelse, NA, NA)
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
edited 2 days ago
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
answered Dec 31 '18 at 22:58
Julius Vainora
32.7k75979
32.7k75979
what is*called as in this context ?
– YOLO
Dec 31 '18 at 23:12
1
@YOLO, I'm not sure what you mean. The first approach is the same aslapply(my_list, function(x) x * NA), so*is the usual product, which results toNAwhen one of the factors isNA.
– Julius Vainora
Dec 31 '18 at 23:15
add a comment |
what is*called as in this context ?
– YOLO
Dec 31 '18 at 23:12
1
@YOLO, I'm not sure what you mean. The first approach is the same aslapply(my_list, function(x) x * NA), so*is the usual product, which results toNAwhen one of the factors isNA.
– Julius Vainora
Dec 31 '18 at 23:15
what is
* called as in this context ?– YOLO
Dec 31 '18 at 23:12
what is
* called as in this context ?– YOLO
Dec 31 '18 at 23:12
1
1
@YOLO, I'm not sure what you mean. The first approach is the same as
lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.– Julius Vainora
Dec 31 '18 at 23:15
@YOLO, I'm not sure what you mean. The first approach is the same as
lapply(my_list, function(x) x * NA), so * is the usual product, which results to NA when one of the factors is NA.– Julius Vainora
Dec 31 '18 at 23:15
add a comment |
7
You can use function is.na<- in a lapply loop.
ret_list <- lapply(my_list, `is.na<-`)
ret_list
#$A
# a b c
#NA NA NA
#
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
add a comment |
7
You can use function is.na<- in a lapply loop.
ret_list <- lapply(my_list, `is.na<-`)
ret_list
#$A
# a b c
#NA NA NA
#
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
add a comment |
7
7
7
You can use function is.na<- in a lapply loop.
ret_list <- lapply(my_list, `is.na<-`)
ret_list
#$A
# a b c
#NA NA NA
#
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
You can use function is.na<- in a lapply loop.
ret_list <- lapply(my_list, `is.na<-`)
ret_list
#$A
# a b c
#NA NA NA
#
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
answered Dec 31 '18 at 23:16
Rui Barradas
16.1k41730
16.1k41730
add a comment |
add a comment |
4
another alternative with dplyr:
lapply(my_list, function(x) dplyr::na_if(x,x))
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
add a comment |
4
another alternative with dplyr:
lapply(my_list, function(x) dplyr::na_if(x,x))
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
add a comment |
4
4
4
another alternative with dplyr:
lapply(my_list, function(x) dplyr::na_if(x,x))
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
another alternative with dplyr:
lapply(my_list, function(x) dplyr::na_if(x,x))
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
answered Dec 31 '18 at 23:36
Mankind_008
1,4762312
1,4762312
add a comment |
add a comment |
4
If the list is not restricted to one level then use rapply.
# test data modified from question
my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))
rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")
which can also be written as:
rapply(my_list2, replace, list = TRUE, values = NA, how = "list")
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
add a comment |
4
If the list is not restricted to one level then use rapply.
# test data modified from question
my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))
rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")
which can also be written as:
rapply(my_list2, replace, list = TRUE, values = NA, how = "list")
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
add a comment |
4
4
4
If the list is not restricted to one level then use rapply.
# test data modified from question
my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))
rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")
which can also be written as:
rapply(my_list2, replace, list = TRUE, values = NA, how = "list")
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
If the list is not restricted to one level then use rapply.
# test data modified from question
my_list2 <- list(list(A = c(a = 1, b = 2, c = 3)), B = c(x = 10, y = 20))
rapply(my_list2, function(x) replace(x, TRUE, NA), how = "list")
which can also be written as:
rapply(my_list2, replace, list = TRUE, values = NA, how = "list")
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
answered Jan 1 at 0:10
G. Grothendieck
145k9126231
145k9126231
add a comment |
add a comment |
3
Another way around
relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)
#$A
# a b c
#NA NA NA
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:30
989
8,91251834
add a comment |
3
Another way around
relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)
#$A
# a b c
#NA NA NA
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:30
989
8,91251834
add a comment |
3
3
3
Another way around
relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)
#$A
# a b c
#NA NA NA
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:30
989
8,91251834
Another way around
relist(replace( unlist(my_list), TRUE, NA ), skeleton = my_list)
#$A
# a b c
#NA NA NA
#$B
# x y
#NA NA
answered Dec 31 '18 at 23:30
989
8,91251834
answered Dec 31 '18 at 23:30
989
8,91251834
answered Dec 31 '18 at 23:30
989
8,91251834
answered Dec 31 '18 at 23:30
989
8,91251834
8,91251834
add a comment |
add a comment |
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1
both approaches seem fine to me (especially if you define the function in Approach 2 beforehand rather than anonymously inline). Why try to be cleverer than that? What specific shortcomings would you like to remedy?
– Ben Bolker
Dec 31 '18 at 23:27
1
Just taking an opportunity to understand R a little better. It seemed to me that this was such a common operation that there had to be a standard, best-practices approach.
– lowndrul
Dec 31 '18 at 23:40